IB QUESTIONS - BONDING

1. Which compound contains both covalent and ionicbonds?

A. sodium carbonate, Na2C03

B. magnesium bromide, MgBr2

e. dichloromethane, CH2CI2D. ethanoic acid, CH3COOH

2. Which pair of elements is most likely to form a covalentlybonded compound?

A. Li and CI

B. P and °e.Ca and S

D. Zn and Br

3. Given the following electronegativities,

H: 2.2 N: 3.0 0: 3.5 F: 4.0which bond would be the most polar?

A. O-H in H20

B. N-F in NF2

e. N-O in N02

D. N-H in NH3

4. What is the correct Lewis structure for methanal?

A. H:C:::O:H e. HC::O:H

D. :C:O:HH

B. HH:C::O:

5. When CH4, NHy H20, are arranged in order of increasingbond angle, what is the correct order?

A. CH4, NHy H20 e. NHy CH4, H20

B. NHy H20, CH4 D. H20, NHy CH4

6. When the H-N-H bond angles in the species NH2, NHyNH: are arranged in order of increasing bond angle(smallest bond angle first), which order is correct?

A. NH2 < NH3 < NH:\ e. NH3 < NH2 < NH:\

B. NH:\ < NH3 < NH2 D. NH2 < NH~ < NH3

7. In which of the following pairs does the second substancehave the lower boiling point?

A. F2, Cl2B. H20, H2S

e.C2H6, C3HS

D. CH30CHy CH3CH20H

8. In which of the following substances would hydrogenbonding be expected to occur?

I. CH4

II. CH3COOH

III. CH30CH3

A. II only

B. I and III only

e. I and III only

D.I, II and III

9. Which one of the following statements is correct?

A. The energy absorbed when liquid ammonia boils isused to overcome the covalent bonds within theammonia molecule.

B. The energy absorbed when sol id phosphorus (P4) meltsis used to overcome the ionic bonds between thephosphorus molecu les.

e. The energy absorbed when sodium chloride dissolvesin water is used to form ions.

D. The energy absorbed when copper metal melts is usedto overcome the non-directional metallic bondsbetween the copper atoms.

10. A solid has a melting point of 1440 "C. It conducts heatand electricity. It does not dissolve in water or in organicsolvents.The bond between the particles is most likely tobe

A. covalent.B. dipole:dipole.

e. ionic.

D. metallic.

11. What are the types of hybridization of the carbon atoms inthe compound

H2CIC-CH2-COOH ?7 2 3

1Sp2

sp

1sp2

sp

12.Which molecule or ion does not have a tetrahedral shape?

A. XeF4

B. SiCl4

e. BF4

D. NH4

13.When the substances below are arranged in order ofincreasing carbon-carbon bond length (shortest bond first),what is the correct order?

I. H2CCH2

A. I < II < III

B. I < III < II

III.©II. H3CCH3

e. II < I < III

D.III < II < I

14. Which of the following species is considered to involve Sp3hybridization?

I. BCI3

A. lonly

B. II only

II. CH4 III. NH3

e. I and III only

D. II and III only

15.How many 11: bonds are present in CO2?

A. One

B. Two

e. Three

D. Four

16. When the following substances are arranged in order ofincreasing melting point (lowest melting point first) thecorrect order is

A. CH3CH2CHy CH3COCHy CH3CH2CH20H

B. CH3CH2CHy CH3CH2CH20H, CH3COCH3e. CHFOCHy CH3CH2CH20H, CH3CH2CH3

D. CH3CH2CH20H, CH3CH2CHy CH3COCH3

IB questions - Bonding 27

-Enthalpy changes

EXOTHERMIC AND ENDOTHERMIC REACTIONSEnergy is defined as the ability to do work, that is, move a force through a distance. It is measured in joules.

Energy = force x distanceW) (N x m)

In a chemical reaction energy is required to break the bonds in the reactants, and energy is given out when new bonds areformed in the products. The most important type of energy in chemistry is heat. If the bonds in the products are stronger thanthe bonds in the reactants then the reaction is said to be exothermic, as heat is given out to the surroundings. Examples ofexothermic processes include combustion and neutralization. In endothermic reactions heat is absorbed from the surroundingsbecause the bonds in the reactants are stronger than the bonds in the products.

The internal energy stored in the reactants is known as its enthalpy,H. The absolute value of the enthalpy of the reactants cannot beknown, nor can the enthalpy of the products, but what can bemeasured is the difference between them, !'!.H. By convention !'!.Hhas a negative value for exothermic reactions and a positive valuefor endothermic reactions. It is normally measured under standardconditions of 1 atm pressure at a temperature of 298 K. Thestandardenthalpy changeof a reaction is denoted by !'!.H"'.

enthalpy, H

reactants

!'!.H = Hproducts - Hreactants(value negative)

products (more stable than reactants)

TEMPERATURE AND HEATIt is important to be able to distinguish between heat andtemperature as the terms are often used loosely.

• Heat is a measure of the total energy in a given amount ofsubstance and therefore depends on the amount ofsubstance present.

• Temperature is a measure of the 'hotness' of a substance.It represents the average kinetic energy of the substance,but is independent of the amount of substance present.

II \r- 50°C ~ r- 50°C ~

Two beakers of water.Both have the sametemperature, but the100 em! of water contains

~ 50 ern!twice as much heat as the

~ 100 cm '50 cm '.

simple calorimeter

polystyrene

reactionmixture

28 Energetics

Representation of exothermic reaction usingan enthalpy diagram.

enthalpy, H

products (less stable than reactants)

!'!.H = Hproducts - Hreactants

(value positive)reactants

Representation of endothermic reaction usingan enthalpy diagram.

CALORIMETRYThe enthalpy change for a reaction can be measuredexperimentally by using a calorimeter. In a simplecalorimeter all the heat evolved in an exothermic reaction isused to raise the temperature of a known mass of water. Forendothermic reactions the heat transferred from the water tothe reaction can be calculated by measuring the lowering oftemperature of a known mass of water.

To compensate for heat lost by the water in exothermicreactions to the surroundings as the reaction proceeds a plotof temperature against time can be drawn. By extrapolatingthe graph, the temperature rise that would have taken placehad the reaction been instantaneous can be calculated.

Compensating for heat lostTo = initial temperature of

reactantsT, = highest temperature

actually reachedT2 = temperature that would have

been reached if no heat:- - - - ___ lost to surroundings, --

T2

~ T,tilCi> !'!.TQ..

E~

To

extrapolation atsame rate ofcooling

!'!. Tfor reaction = T2 - To

reactants mixed

time

AH calculationsCALCULATION OF ENTHALPY CHANGESThe heat involved in changing the temperature of any substance can be calculated from the equation:

Heat energy = mass (m) x specific heat capacity (c) x temperature change (t- 7)

The specific heat capacity of water is 4.18 kJkg-1 K-l. That is, it requires 4.18 kilojoules of energy to raise the temperature ofone kilogram of water by one kelvin.

Enthalpy changes are normally quoted in k] mol-1, for either a reactant or a product, so it is also necessary to work out thenumber of moles involved in the reaction which produces the heat change in the water.

Worked example 150.0 cm ' of 1.00 mol drrr ' hydrochloric acid solution wasadded to 50.0 cm' of 1.00 mol drrr-' sodium hydroxidesolution in a polystyrene beaker. The initial temperature ofboth solutions was 16.7 "C. After stirring and accounting forheat loss the highest temperature reached was 23.5 "C.Calculate the enthalpy change for this reaction.

Step 1. Write equation for reaction

HCI(aq) + NaOH(aq) -;> NaCI(aq) + H20(I)

Step 2. Calculate molar quantities

Amount of HCI = ;go~ x 1.00 = 5.00 x 10-2 mol

Amount of NaOH = 15go~x 1.00 = 5.00 x 10-2 mol

Therefore the heat evolved will be for 5.00 x 10-2 mol

Step 3. Calculate heat evolved

Total volume of solution = 50.0 + 50.0 = 100 cm3

Assume the solution has the same density and specificheat capacity as water thenmass of 'water' = 100 g = 0.100 kgTemperature change = 23.5 -16.7 = 6.8 °C = 6.8 KHeat evolved in reaction = 0.100 x 4.18 x 6.8 = 2.84 kJ

= 2.84 kJ (for 5.00 x 10-2 mol)

t-H for reaction = - 2.84 x 5.001x 10 2= -56.8 kJ mor '

(negative value as the reaction is exothermic)

Worked example 2A student used a simple calorimeter to determine theenthalpy change for the combustion of ethanol.

C2H50H(I) + 302(g) -;> 2C02(g) + 3Hp(l)

Worked example 350.0 ern- of 0.200 mol drrr-' coppenlt) sulfate solution was placed in a polystyrene cup. After two minutes 1.20 g of powderedzinc was added. The temperature was taken every 30 seconds and the following graph obtained. Calculate the enthalpy changefor the reaction taking place.

beaker-_++-_ __ thermometerI-f-----

f-I- r--t-- r----- 250g of

water

~ ~ j..------spirit lamp

~ j-----ethanol-~~~~. \

air hole

clamp-

draughtshield ---

When 0.690 g (0.015 mol) of ethanol was burned itproduced a temperature rise of 13.2 K in 250 g of water.Calculate t-H for the reaction.

Heat evolved = 250 x 4.18 x 13.2 = 13.79 kJby 0.015 mol 1000

1t-H=-13.79x 0.015=-920 kJ mor '

Note: the Data Book value is -1371 k] mol:". Reasonsforthe discrepancy include the fact that not all the heatproduced is transferred to the water, the water losessomeheat to the surroundings, and there is incompletecombustion of the ethanol.

29

'-....zincadded

Step 1. Write the equation for the reaction

Cu2+(aq) + Zn(s) -;> Cu(s) + Zn2+(aq)

Step 2. Determine the limiting reagent

Amount of Cu2+(aq) = 50.0/1000 x 0.200 = 0.0100 mol

Amount of Zn(s) = 1.20/65.37 = 0.0184 mol

:. Cu2+(aq) is the limiting reagent

Step 3. Extrapolate the graph (already done) to compensate for heat lossand determine t- T

st: 10.4°C

- -- - -27 - - - --- -

~ 25•.•.•.

~::J 23-::0Q;~ 212

t-T

19

17

Step 4. Calculate the heat evolved in the experiment for 0.0100 mol ofreactants

Heat evolved = 50.0/1000 x 4.18 x 10.4 °C = 2.17 kJtime/ min

Step 5. Expressthis as the enthalpy change for the reaction t-H = -2.17 x _1_ = -217 kJ mor '0.0100

Energetics 29

Bond enthalpies and Hess' lawBOND ENTHALPIESEnthalpy changes can also be calculated directly from bond enthalpies.For a diatomic molecule the bond enthalpy is defined as the enthalpy changefor the process

X-Y(g) = X(g) + Y(g) Note the gaseous state.

For bond formation the value is negative and for bond breaking the value ispositive. If the bond enthalpy values are known for all the bonds in thereactants and products then the overall enthalpy change can be calculated.

Some average bond enthalpiesAll values in kJ mcr '

H-H 436 C=C 612 C=C 837C-C 348 0=0 496 N=N 944C-H 412O-H 463N-H 388N-N 163

Worked example 1Hydrogenation of ethene

Worked example 2Combustion of hydrazine in oxygen (this reaction has beenused to power spacecraft)

+H HI I

H-H (g) ~ H-y-C(- H (g)H H

energy released whenbonds are formed:

H H'N-N/ (g) + 0=0 (g) ~

H/ 'H

energy absorbed:

°N=N (g) + 2 / <, (g)H Henergy absorbed tobreak bonds: energy released:

C=C 612 }4 C-H 4 x 412 2696 kJ

H-H 436

C-C 348 }6 C-H 6 x 412 2820 kJ

N-N 163} N=N 944 }4 N-H 4 x 388 2211 4 O-H 4 x 463 2796

0=0 496

!::.H = -(2796 - 2211) = -585 kJ rnor 'There is more energy released than absorbed so the reactionis exothermic.

!::.H = -(2820 - 2696) = -124 kJ rnor '

LIMITATIONS OF USING AVERAGE BOND ENTHALPIESAverage bond enthalpies can only be used if all the reactants and products are in the gaseous state. If water were a liquidproduct in the above example then even more heat would be evolved since the enthalpy change of vaporization of water wouldalso need to be included in the calculation.

In the above calculations average bond enthalpies have been used. These have been obtained by considering a number ofsimilar compounds. In practice the energy of a particular bond will vary slightly in different compounds, as it will be affected byneighbouring atoms. So!::.H values obtained from using average bond enthalpies will not necessarily be very accurate.

HESS'LAWHess' law states that the enthalpy change for a reaction depends only onthe difference between the enthalpy of the products and the enthalpy of thereactants. It is independent of the reaction pathway.

The enthalpy change going !::.H2from A to B is the same whether A ----> C

!::.H1 1 /!::.H3 !::.H1 = !::.H2 + !::.H3the reaction proceeds directlyto A or whether it goesvia an intermediate. B

This law is a statement of the law of conservation of energy. It can be usedto determine enthalpy changes, which cannot be measured directly. Forexample, the enthalpy of combustion of both carbon and carbon monoxideto form carbon dioxide can easily be measured directly, but the combustionof carbon to carbon monoxide cannot. This can be represented by anenergy cycle.

-393 = !::.Hx + (-283)

!::.Hx = -393 + 283 = -110 kJ mor '-393kJ mor '

30 Energetics

••

Worked example 3Calculate the standard enthalpy change whenone mole of methane is formed from itselements in their standard states. The standardenthalpies of combustion !::.H~ of carbon,hydrogen, and methane are -393, -286, and-890 k] mol " respectively.

Step 1. Write the equation for the enthalpychange with the unknown !::.H" value. Call thisvalue !::.H: .

!::.H"C(s) + 2H2(g) ~ CH4(g)

Step 2. Construct an energy cycle showing thedifferent routes to the products (in this case theproducts of combustion)

!::.H"C(s) + 2H2(g) x , CH4(g)

102(g) 102(g) ~g)

CO2(g) + 2H2°(l)

Step 3. Use Hess' law to equate the energychanges for the two different routes

!::.H; (C) + 2!::.Hc" (H2) !::.H; + !::.H; (CH4)

direct route route via methane

Step 4. Rearrange the equation and substitutethe values to give the answer

!::.Hx"= !::.H; (C) + 2!::.H; (H2) - !::.H; (CH4)

= -393 + (2 x -286) - (-890) kJ rnor '

= -75 kJ rnor '

Energy cycles

STANDARD ENTHALPY CHANGE OF FORMATION I1H;The standard enthalpy change of formation of a compound is the enthalpy change when one mole of the compound is formedfrom its elements in their standard states at 298 K and 1 atm pressure. From this it follows that tlHr for an element in itsstandard state will be zero. The standard enthalpy change of combution, tlH:' is the enthalpy change when one mole of asubstance is completely com busted in oxygen under standard conditions, An accurate value for the standard enthalpy changeof formation of ethanol can be determined from the following cycle.

tlrf (C2H50H)

+ 3H2(g) )~~02(9) C2H50H(I)

3 x tlHr (H20) j 1~ 02(g) 302(g)

H~(C2H50H)+ 3H20(I)

2C(s)

2 x A,,· (CO,120,(9)

2C02(g)

By Hess' law: tlH/'(C2H50H) = 2 x tlH/'(C02) + 3 x tlH/"(H20) - tlHC&(C2H50H)

Substituting the relevant values tlH/'(C2H50H) = (2 x -393.5) + (3 x -285.8) - (-1371) = -273.4 kJ rnor '

'ili'ORN:HABERCYCLES-Born-Haber cycles are simply energy cycles for the formation ofionic compounds. The enthalpy change of formation of sodium chlo-ride can be considered to occur through a number of separate steps.

Na(s) +

1 tlH~(Na)Na(g)

1 tlHI~(Na)Na+(g) +

Ml&(NaCI)~C12 f ,Na+CI-(s)

1 tlH~(CI)CI(g)

1 tlH~A(CI) tlH1:tt(NaCI)

CI-(g)

Using Hess' law:

tlH/,(NaCI) = tlH;(Na) + tlH1E(Na) + tlH;(CI) + tlH~A(CI)+ tl~att(NaCI)

Substituting the relevant values:tlHr(NaCI) = +108 + 494 + 121 - 364 -771 = -412 kJ mor '

Note: it is the large lattice enthalpy that mainly compensates for theendothermic processes and leads to the enthalpy of formation ofionic compounds having a negative value.

LA TTICE ENTHALPY I1Ht:ttThe lattice enthalpy relates either to the endothermic process ofturning a crystalline solid into its gaseous ions or to the exothermicprocess of turning gaseous ions into a crystalline solid.

MX(s) = M+(g) + X-(g)

The sign of the lattice enthalpy indicates whether the lattice is beingformed (-) or broken (+).

The size of the lattice enthalpy depends both on the size of the ionsand on the charge carried by the ions.

anion size increasing

NaCI NaBr NaI

cation size increasing

LiCI NaCI KCI

Lattice enthalpy / kJ mol-1

846 771 701 771 733 684

charge on anion increasing

MgCI2 MgO

charge on cation increasing

NaCI MgCI2

Lattice enthalpy / k] mol:"

771 2493 2493 3889

ENTHALPY OF A TOMIZA T;ON I1H;;= ."The standard enthalpy of atomization is the standardenthalpy change when one mole of gaseous atoms isformed from the element in its standard state understandard conditions. For diatomic molecules this isequal to half the bond dissociation enthalpy.

~ CI2(g) ~ CI(g) tlH; = + 121 kJ mor '

ELECTRON AFFINITY I1HE~The electron affinity is the enthalpy change when anelectron is added to an isolated atom in the gaseousstate, i.e.

Atoms 'want' an extra electron so electron affinityvalues are negative for the first electron. However,when oxygen forms the 02- ion the overall process isendothermic:

tlH& = -142 kJ rnot '

O-(g) + e- ~ 02-(g) tlH& = +844 kJ mol-1

overall O(g) + 2e- ~ 02-(g) tlH& = +702 kJ mor '

USE OF BORN-HABER CYCLESLike any energy cycle Born-Haber cycles can be usedto find the value of an unknown. They can also beused to assess how ionic a substance is. The latticeenthalpy can be calculated theoretically byconsidering the charge and size of the constituent ions.It can also be obtained indirectly from the Born-Habercycle. If there is good agreement between the twovalues then it is reasonable to assume that there is ahigh degree of ionic character, e.g. NaCI. However, ifthere is a big difference between the two values then itis because the compounds possesses a considerabledegree of covalent character, e.g. AgCl.

NaCI AgClTheoretical value / kJ mol:" 766 770Experimental value / kJ mol-1 771 905

Energetics 31

Entropy and spontaneity

DISORDERIn nature systems naturally tend towards disorder. Anincrease in disorder can result from:

• mixing different types of particles, e.g. the dissolving ofsugar in water

• a change in state where the distance between theparticles increases, e.g. liquid water -'? steam

• the increased movement of particles, e.g. heating a liquidor gas

• increasing the number of particles, e.g.2HP2(1) -'? 2H20(I) + 02(g)·

The greatest increase in disorder is usually found where thenumber of particles in the gaseous state increases.

The change in the disorder of a system is known as theentropy change, t,S.The more disordered the systembecomes the more positive the value of t,S becomes.Systems which become more ordered will have negativet,S values.

(two molesof gas)

(one moleof solid)

SPONTANEITYA reaction is said to be spontaneous if it causes a systemto move from a less stable to a more stable state. This willdepend both upon the enthalpy change and the entropychange. These two factors can be combined andexpressed as the Gibbs energy change t,C, often knownas the 'free energy change'.

The standard free energy change t,C& is defined as:

t,G& = t,H& - Tt,S&

Where all the values are measured under standardconditions. For a reaction to be spontaneous it must beable to do work, that is t,C& must have a negative value.

Note: the fact that a reaction is spontaneous does notnecessarily mean that it will proceed without any input ofenergy. For example, the combustion of coal is aspontaneous reaction and yet coal is stable in air. It willonly burn on its own accord after it has received someinitial energy so that some of the molecules have thenecessary activation energy for the reaction to occur.

32 Energetics

ABSOLUTE ENTROPY VALUESThe standard entropy of a substance is the entropy changeper mole that results from heating the substance from 0 Kto the standard temperature of 298 K. Unlike enthalpy,absolute values of entropy can be measured. The standardentropy change for a reaction can then be determined bycalculating the difference between the entropy of theproducts and the reactants.

ss: = S& (products) - S& (reactants)

e.g. for the formation of ammonia

3H2(g) + N2(g) ¢ 2NH3(g)

the standard entropies of hydrogen, nitrogen, and ammoniaare respectively 131,192, and 192 J K-l mol ".

Therefore per mole of reaction

t,S& = 2 x 192 - [(3 x 131) + 192] = -201 J K-lmol-1

I f . & -201 1 1(or per rno e 0 ammonia t,S = -2- = -101 J K- mol- )

Spontaneity of a reaction

POSSIBLE COMBINATIONS FOR FREEENERGY CHANGESSome reactions will always be spontaneous. If f);Hf> isnegative or zero and f);Sf> is positive then f);Cf> mustalways have a negative value. Conversely if f);Hf> ispositive or zero and f);Sf> is negative then f);Cf> mustalways be positive and the reaction will never bespontaneous.

For some reactions whether or not they will be spontaneousdepends upon the temperature. If f);Hf> is positive or zeroand f);Sf> is positive, then f);Cf> will only become negativeat high temperatures when the value of Tf);Sf> exceeds thevalue of f);Hf> .

Type f);Hf> f);Sf> tss: f);Hf> _ Tf);Sf> f);Cf>

1 0 + + (0) - (+)2 0 (0) - (-) +3 + + (-)-(+)

4 + (+) - (-) +5 + + + (+)-(+) -or +6 (-)-(-) + or-

Type 1. Mixing two gases. f);Cf> is negative so gases willmix of their own accord. Gases do not unmix oftheir own accord (Type 2) as f);Cf> is positive.

Type 3. (NH4)2Cr207(S) -- N2(g) + Cr203(s) + 4H20(g)

The decomposition of ammonium dichromate isspontaneous at all temperatures.

Type 4. N2(g) + 2H2(g) --> N2H4(g)

The formation of hydrazine from its elements willnever be spontaneous.

Type 5. CaC03(s) -- CaO(s) + CO2 (g)

The decomposition of calcium carbonate is onlyspontaneous at high temperatures.

Type 6. C2H4(g) + H2(g) --> C2H6(g)

Above a certain temperature this reaction willcease to be spontaneous.

~~~ ••• :f •• r.~~~~~DETERMINING THE VALUE OF LlG"The precise value of f);Cf> for a reaction can be determinedfrom f);C~ values using an energy cycle, e.g. to find thestandard free energy of combustion of methane given thestandard free energies of formation of methane, carbondioxide, water, and oxygen.

f);G:

CH4(g) + 202 -- CO2(g) + 2H20(I)

~/C(s) + 202(g) + 2H2(g)

By Hess' law

f);G: = [f);G: (C02) + 2f);G: (HP)]- [f);G: (CH4) +2f);Gt(02)]

Substituting the actual values

f);G: = [-394 + 2 x (-237)]- [-50 + 2 x 0] = -818 kJ mor '

sc: values can also be calculated from using the equationsc: = f);Hf> - tss»,For examole, in Type 5 in the adjacentlist the values for f);Hf> and ss: for the thermaldecomposition of calcium carbonate are + 178 k] mol-1 and+ 165.3 J K-1 mol"! respectively. Note that the units of f);Sf>are different to those of f);Hf> .

At 25 DC (298 K) the value for f);Gf> = 178 - 298 x 165.31000

= +129 kJ mor '

which means that the reaction is not spontaneous.

The reaction will become spontaneous when Tf);S" > f);Hf>.

nSf> = f);H" when T = f);Hf>= 178 1077 K (804 DC)f);Sf> 165.3/1000

Therefore above 804 DCthe reaction will be spontaneous.

Note: this calculation assumes that the entropy value isindependent of temperature, which is not strictly true.

18 QUESTIONS - ENERGETICS 3.

products

1. Which statement about this reaction is correct?

2Fe(s) + 3C02(g) --> Fe203(s) + 3CO(g) f);Hf> = +26.6 k]

A. 26.6 kJ of energy are released for every mole of Fereacted

B. 26.6 k] of energy are absorbed for every mole of Fereacted

C. 53.2 k] of energy are released for every mole of Fereacted

D. 13.3 k] of energy are absorbed for every mole of Fereacted

2. When solutions of HCI and NaOH are mixed thetemperature increases. The reaction:

W(aq) + Ol-ltaq) --> H20(I)

A. is endothermic with a positive f);Hf>.

B. is endothermic with a negative f);Hf>.

C. is exothermic with a positive f);Hf>.

D. is exothermic with a negative f);Hf> .

reactants

What can be deduced about the relative stability of thereactants and products and the sign of f);Hf> r from theenthalpy level diagram above?

Relative stability Sign of f);Hf>A. Products more stable

B. Products more stable +

C. Reactants more stable

D. Reactants more stable +

18 questions - Energetics 33

4. For the reaction:

2C(s) + 2H2(g) ---;.C2H4(g) llH~ = +52.3 k]

If llH~ = -174.4 k] for the reaction:

C2H2(g) + H2(g) ---;.C2H4(g)

what can be said about the value of llH~ for the reactionbelow?

2((s) + H2(g) ---;. C2H2(g)

A. llH~ must be negative.

B. llH~ must be a positive number smaller than 52.3 k].C. llH~ must be a positive number larger than 52.3 k].

D. No conclusion can be made about llH~ without thevalue of H for H2(g).

5. The enthalpy changes for two different hydrogenationreactions of C2H2 are:

C2H2 + H2 ---;.C2H4 llH~

C2H2 + 2H2 ---;.C2H6 llH~

Which expression represents the enthalpy change for thereaction below?

C2H4 + H2 ---;.C2H6 llH" = ?

A. llH~ + llHr

B. llHf - llHr

c. llH~ - llHf

D. -IlHf - llH~

6. 2KHC03(s) llH" , K2C03(S) + cO2 (g) + H20(l)

+2HCI(aq~ llH,{/ +2HCI(aq)=, /2KCI(aq) + 2C02(g) + 2H20(I)

This cycle may be used to determine llH" for thedecomposition of potassium hydrogencarbonate.Which expression can be used to calculate llH"?

A. llH" = llHf + llH~

B. llH" = llHf - llH'{

c. llH" = ~IlHf - llHr

D. llH" = llH~ - llHf

7. Use the bond energies for H-H (436 k] mol:").Br-Br (193 kJ rnol ") and H-Br (366 kJ mol:") tocalculate llH" (in k] mol ") for the reaction:

A. 263

B. 103

C.-103

D. -263

8. The average bond enthalpy for the C-H bond is412 k] mol". Which process has an enthalpy changeclosest to this value?

A. CH4(g) ---;.Cis) + 2H2(g)

B. CH4(g) ---;.Ctg) + 2H2(g)

c. CH4(g) ---;.Cis) + 4H(g)

D. CH4(g) ---;.CH3(g) + H(g)

9. Which of the changes below occurs with the greatestincrease in entropy?

A. Na20(s) + H20(I) ---;.2Na+(aq) + 20H-(aq)

B. NH3(g) + HCI(g) ---;.NH4C1(s)

C. H2(g) + I2(g) ....,.2HI(g)

D. Cis) + CO2 (g) ....,.2CO(g)

10. How would this reaction at 298 K be described inthermodynamic terms?

2H20(g)""" 2H2(g) + °2(g)

A. Endothermic with a significant increase in entropy

B. Endothermic with a significant decrease in entropy

C. Exothermic with a significant increase in entropy

D. Exothermic with a significant decrease in entropy

11. The enthalpy change, llH", for a chemical reaction is-10 k] mor' and the entropy change, 115", is -10 JK-l rnol!at 27°C. What is the value of sc: (in J) for this reaction?

A. -260

B. -7000

C. -9730

D. -13000

34 IB questions - Energetics (contd)

12. At 0 °C, the mixture formed when the following reactionreaches equilibrium consists mostly of N204(g).

2N02(g)= NP4(g)

What are the signs of llC, llH, 115 at this temperature?

llG llH IlSA. + + +

B.C. + +

D. + +

13. Which substance has the largest lattice energy?

A. NaF

B. KCI

C. MgO

D. CaS

14. Which factor(s) will cause the lattice enthalpy of ioniccompounds to increase in magnitude?

I. an increase in the charge on the ions

II. an increase in the size of ions

A. lonlyB. II only

C. Both I and II

D. Neither I nor II

15. The Born-Haber cycle for the formation of potassiumchloride includes the steps below:

I. K(g)""" K+(g) + e-

ll. ICI2(g)""" CI(g)

III. CI(g) + e....,.CI-(g)

IV. K+(g) + CI-(g) ....,.KCI(s)

Which of these steps are exothermic?

A. I and II only

B. III and IV only

C. I, II and II only

D. I, III and IV only